& Tornado Alley
That Would Be Created On
a Rotating Spherical Planet
Because of a planet rotation, the fluids on its surface would move to its equator unless a countervailing pressure gradient is established. Below is given a derivation of the pressure distribution on the surface of a rotating spherical planet which would be required to maintain surface equilibrium.
Let r be the distance to the center of the planet and Ω its rate of rotation. The latitude angle is denoted φ and the longitude θ At a latitude φ the distance to the axis of the planet is rcos(φ). The centrifugal force on a particle of mass dm at latitude φ and distance to the planet's center of r is
This may be resolved into a radial and a latitudinal component; i.e.,
The mass in an infinitesimal volume dV is ρdV where ρ is the fluid density. An infinitesimal volume between r and r+dr, φ and φ+dφ, and θ and θ+dθ is r²cos(φ)drdφθ. This can be represented as an area times a thickness in three ways:
The force on the infinitesimal element due the pressure p on the face at φ+dφ less the force on face at φ is
When this force is added to the latitudinal component of the centrifugal force on dV the result must be zero; i.e.,
Integrating this last equation with respect to φ from 0 to Φ with r constant gives:
Therefore the equilibrium latitudinal pressure distribution is given by
The area of the top and bottom surfaces of the infinitesimal volume are approximately equal to dB = r²cos(φ)dφθ. The infinitesimal volume is then dV=dBdr.
The difference in pressure on the top and bottom surfaces plus the gravitational pull on the infinitesimal volume plus the centrifugal force must equal zero; i.e.,
When this last equation is integrated with respect to r from R0 to R the result is:
This result presumed the gravitational attraction is constant with altitude, R−R0. More generally the result would be the pressure as a function of altitude with a correction for the effect of the centrifugal force. The correction term depends upon the cosine squared of the latitude.
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